Let S = \ (x, y) ∈ R ^ 2 : y ^ 2 1 + r - x ^ 2 1 - r = 1\ , where r ≠ ± 1 . Then S represents:

Correct answer: C. An ellipse whose eccentricity is (2)/(r + 1) , when r > 1 Given y ^ 2 1 + r - x ^ 2 1 - r = 1 If r ∈ (0, 1) then this curve is a hyperbola whose eccentricity is 1 + (1 - r)/(1 + r) = (2)/(r + 1) . If r > 1 , then this curve is an ellipse whose eccentricity is 1 - (r - 1)/(r + 1) = (2)/(r + 1) .

JEE Main 201910 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2019 · 10 January · Evening Shift · Mathematics · Ellipse

Let S = \ (x, y) ∈ R ^ 2 : y ^ 2 1 + r - x ^ 2 1 - r = 1\ , where r...

Let

S = \left\{\left(x, y\right) \in {R}^{2}: \dfrac{{y}^{2}}{1 + r} - \dfrac{{x}^{2}}{1 - r} = 1\right\},

where

r \ne \pm 1 .

Then

S

represents:

Options

AAn ellipse whose eccentricity is $\dfrac{1}{\sqrt{r + 1}},$ when $r > 1 .$
BA hyperbola whose eccentricity is $\dfrac{2}{\sqrt{r + 1}},$ when $0 < r < 1 .$
CAn ellipse whose eccentricity is $\sqrt{\dfrac{2}{r + 1}},$ when $r > 1$
DA hyperbola whose eccentricity is $\dfrac{2}{\sqrt{1 - r}},$ when $0 < r < 1$

Correct answer

C. An ellipse whose eccentricity is $\sqrt{\dfrac{2}{r + 1}},$ when $r > 1$

Step-by-step solution

Given

\dfrac{{y}^{2}}{1 + r} - \dfrac{{x}^{2}}{1 - r} = 1

r \in \left(0, 1\right) then this curve is a hyperbola whose eccentricity is

\sqrt{1 + \dfrac{1 - r}{1 + r}} = \sqrt{\dfrac{2}{r + 1}}

. <br /><br />If

r > 1

, then this curve is an ellipse whose<br />eccentricity is

\sqrt{1 - \dfrac{r - 1}{r + 1}} = \sqrt{\dfrac{2}{r + 1}}$.

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